Question

If a curve is given by $x=a\cos t+\frac{b}{2}\cos 2t$ and $y=\sin t+\frac{b}{2}\sin 2t$, then the points for which $\frac{d^{2}y}{d{x}^2}=0$, are given by.

• A. $\sin t=\frac{2a^2+b^2}{3ab}$
• B. $\cos t=-\left[\frac{a^2+2b^2}{3ab}\right]$
• C. $\tan t=a/b$
• D. None of the above

Answer 1

Answer: (B)

$\cos t=-\left[\frac{a^2+2b^2}{3ab}\right]$

We have,
$x=a\cos t+\frac{b}{2}\cos 2t$
and $y=a\sin t+\frac{b}{2}\sin 2t$
On differentiating both equations w.r.t. t, we get
$\frac{dx}{dt}=-a\sin t-b\sin 2t$
and $\frac{dy}{dt}=a\cos t+b\cos 2t$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t+b\cos 2t}{-a\sin t-b\sin 2t}$
$\Rightarrow \frac{dy}{dx}=-\frac{(a\cos t+b\cos 2t)}{(a\sin t+b\sin 2t)}$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{d}{dt}\left(\frac{a\cos t+b\cos 2t}{a\sin t+b\sin 2t}\right)\frac{dt}{dx}$
$=\frac{\left[\begin{array}{c}(a\sin t+b]sin2t\left)\right(-a\sin t-2b\sin 2t)\\ -(a\cos t+b\cos 2t)(a\cos t+2b\cos 2t)\end{array}\right]}{(a\sin t+b\sin 2t{)}^2}\times \frac{1}{-a\sin t-b\sin 2t}$
$=\frac{\left[\begin{array}{c}{a}^2{\sin }^{2}t+3ab\sin t\sin 2t+2b^2{\sin }^{2}2t\\ +a^2{\cos }^{2}t+3ab\cos t\cos 2t+2b^2{\cos }^{2}2t\end{array}\right]}{(a\sin t+b\sin 2t{)}^3}$
$=\frac{a^2({\sin }^{2}t+{\cos }^{2}t)+b^2({\sin }^{2}2t+{\cos }^{2}2t)+3ab\cos (2t-t)}{(a\sin t+b\sin 2t{)}^3}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left[\frac{a^2+2b^2+3ab\cos t}{(a\sin t+b\sin 2t{)}^3}\right]$
Given, $\frac{d^{2}y}{d{x}^2}=0$
$\Rightarrow a^2+2b^2+3ab\cos t=0$
$\Rightarrow \cos t=-\left(\frac{a^2+2b^2}{3ab}\right)$